Closed in what topology? I tend to view B as a closed subset of B ...
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Yemon ChoiNov 17 '10 at 2:08

Yemon, in light of the question linked to by David, the OP is likely thinking of having A and B in some other background space, rather than the subspace topology on $B$.
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Joel David HamkinsNov 17 '10 at 2:26

Joel: that was my reading too, but I was hoping that the OP might come back and clarify (this tending to help in answering one's own questions, or getting other people to answer them).
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Yemon ChoiNov 17 '10 at 4:13

1 Answer
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The answer to your first question is no. Let $A$ be a sequence on the $x$ axis converging to the origin in the real plane $\mathbb{R}^2$, which is the background topological space for these examples. Let $B$ be a cone over the sequence, connecting each point in the sequence to the point $(0,1)$, together with the origin, but not connecting the origin to $(0,1)$. Then $B$ is connected, but not closed, and there is no connected closed set $C$ contained in $B$ and containing $A$, since any connected set $C$ contained in $B$ and containing $A$ would have to contain all the lines and thus be all of $B$, which is not closed in the plane.

The example can be extended to an answer for your path-connected question as well. We simply add a roundabout path from the origin to $(0,1)$. That is, $A$ is the sequence $(\frac{1}{n},0)$ plus $(0,0)$ in the real plane. The set $B$ contains lines from every $(\frac{1}{n},0)$ to $(0,1)$, and $B$ also includes a circular curve connecting $(0,0)$ to $(1,0)$ through the left half-plane. So $A$ is closed, $B$ is path-connected, but any closed path-connected set $C$ with $A\subset C\subset B$ will have to contain all those lines and hence $C=B$, but this is not closed in the plane.

Similarly, for your last question, any closed connected $C$ with $A\subset C\subset B$ for the same example will have to contain all the lines from the points on the sequence to $(0,1)$, and hence not be closed (since it will miss the line joining the origin to $(0,1)$, which is not in $B$.